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Homework Help: Cosmology question

  1. Mar 16, 2008 #1
    Hi guys, I'm in a pickle here with an assignment.

    I'm required to find the redshift at which the energy density in radiation equals the energy density in matter.
    The following are the givens we were...well, given:
    1) assume 1 baryon = 10^9 photon
    2) assume all photons have energy to the wavelength of the peak of a 2.73K black-body radiation curve
    3) scale factor (a) of matter: Pm = a^-3
    4) scale factor of radiation: Pr = a^-4
    5) we are to work out the energy density of a photon & baryon in present time and then apply the scale factor

    Apparently this isn't using complicated formulas, just simple E=mc^2 & E=h*nu.

    me right now = lost...need major help

    Thanks in advance
  2. jcsd
  3. Mar 16, 2008 #2


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    I don't know what assumption #1 means exactly

    but suppose it gives you a way of comparing today's radiation energy density with today's matter density.

    Just to use different numbers, suppose that TODAY the CMB radiation density is one MILLIONTH of the matter density (expressed as energy equivalent)

    then that means if you go back to redshift z+1 = 1,000,000 the densities will be equal

    Because if you go back to redshift z+1 = N
    then the radiation density will be N4 times what it is today
    and the matter density will be N3 times what it is today
    so the ratio of radiation to matter densities will increase a factor of N.

    If matter, today, has an advantage of a factor of N, then you have to go back to redshift z+1 = N in order to bring them into balance

    so the problem is easy once you know what the ratio is today. (of CMB radiation to matter)

    the other kinds of radiation are minor compared to CMB so can forget about them
  4. Mar 16, 2008 #3
    well, the prof wanted us to solve the redshift using the photon-baryon ratio, hence the first assumption...and combined w/ your explanation marcus, i'm deducing that i can use E=mc2 to work out the energy density for matter & use E=hv to work out the energy density for radiation...which will give me the radiation-matter ratio, then apply the scale factor.

    would that be correct?
  5. Mar 17, 2008 #4
    im trying to do this problem too dum-dum so when you get an answer send me an email and we can compare answers(sndhooper@hotmail.com)

    and marcus #1 just means that there are and always have been approximately 10^9 photons for every baryon in the universe
  6. Mar 17, 2008 #5


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    The baryon/photon ratio remains approximately constant over between matter/radiation equality and now. So i) compute the energy in a 2.73K CMB photon and compare the energy in 10^9 of them with that of a single baryon (use E=mc^2). If you change the scale factor a by a factor of 1/N, each photon gains in energy by a factor of N and the baryon remains unchanged. This is because the energy density (NOT scale factor) of matter goes like 1/a^3 and that of radiation by 1/a^4. Now just figure out what change in scale factor a will render them approximately equal.
  7. Mar 18, 2008 #6
    Hey dick, about the E=mc^2 would you plug in 2.73 and 10^9 into E since mass is not given and c is speed of light. So we need to find m then? Sorry I'm not great with math especially physics lol.
  8. Mar 18, 2008 #7
    we are to use the mass of a proton for 'm'...

    you don't plug 109 into anything, you need to multiple that to the energy density of a photon (representing the radiation) to make it on par w/ the energy density of a baryon (representing the matter)
  9. Mar 18, 2008 #8
    I see ok so you find the proton mass which represent the baryon mass. I calculated the photon energy using E=hv is that right? since v=1.9 when it is at 2.7K. What does "a" mean in a^-3. So what do you do with all the numbers you find? do you sub it with a?
  10. Mar 18, 2008 #9
    hmmm just a quick question like we know v=1.9nm so in order to like match up with proton SI units, do we need to convert the 1.9nm into meters and then do the E=hv? Because I got a 10^-42 and it looks awfully small! thnx again

    this is my email harvey_shadowfalcon@hotmail.com I think discussing this over msn is better
    Last edited: Mar 18, 2008
  11. Mar 18, 2008 #10
    so is the energy density of photon is E=(6.63 10^-34)(1.9 10^-9)?
  12. Mar 18, 2008 #11
    use the E=hv equation for energy density of photon (meaning radiation)
    h = planck's constant
    v = frequency (you're giving wavelength, so freq = spd of light/wavelength)
  13. Mar 18, 2008 #12
    can you explain a bit more what i do after i find the ratio between photons and baryons. this whole scale factor thing is a bit confusing
  14. Mar 19, 2008 #13
    look at marcus's explanation...it's about as thorough as i can think of as well.
  15. Mar 19, 2008 #14
    so the ratio of Ephoton/Ebaryon = n?
  16. Mar 19, 2008 #15
    is anyone else getting 2.414x10^11 baryons : 1 photon ?
  17. Mar 19, 2008 #16
    umm, i don't think that's right mixx.

    we're trying to solve for "z"...
  18. Mar 19, 2008 #17
  19. Mar 19, 2008 #18
    woe is me. i just redid my calculations... im getting my ratio as 7.24 x 10^19 baryons:photon
  20. Mar 19, 2008 #19


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    this is clearly a homework so I cant just say the answer and you all seem well on the way.

    but I will describe a DIFFERENT WAY TO DO IT.

    You can get the energy density of CMB radiation in terms of joules per cubic meter
    using a need trick which I will show you (the fourth power law)

    And the energy density of matter (dark and baryonic combined) is 0.2 joule per cubic kilometer, which is 0.2 x 10-9 joule per cubic meter. So once you have the two energy densities it is easy to take the ratio----which is essentially the same as the redshift.

    It is a blackbody radiation law that if the temperature is T then the density of radiation is arad T4

    where arad is called the RADIATION CONSTANT

    Many times people simply use the symbol "a" for the radiation constant but we already have that letter used in this problem for the scale factor, so to avoid confusion lets call it arad , like I said.

    So it is really trivial. You just take the temperature 2.7 kelvin and raise to fourth power and multiply by arad and that will give you some small quantity of joule per cubic meter. then you take the ratio, and you are done.

    But be warned, this does not seem to be the way the teacher wants you to solve the problem. So don't get sidetracked by my method. It is just for comparison.
    Last edited: Mar 19, 2008
  21. Mar 19, 2008 #20


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    the radiation constant is 7.566 x 10-17 joules per cubic meter per kelvin4

    and I guess 2.73 kelvin raised to the fourth power is roughly 55 kelvin4

    so multiply that by the radiation constant and you get roughly 42 x 10-16 joules per cubic meter

    that is 4.2 millionths of a joule per cubic kilometer. that is the energy density of the CMB.

    and we already said that the energy density of all types matter (baryonic + dark) was 0.2 joules per cubic kilometer. that is 200,000 millionths

    so as a rough calculation the ratio is about 200,000/4.2 which is about 50,000

    I wonder if that is approximately right. Or have I made a mistake.

    In any case the professor won't like it if you solve it this way, even if the answer is right. Because he did not say to use the radiation constant.
    It makes a huge difference whether you count dark matter or not. This professor seems only to be concerned with baryonic matter
    which I estimate has a presentday density of 34,000 millionths of a joule per cubic kilometer. And the CMB density is currently 4.2 millionths. So the ratio is only 34000/4.2 = 8100.
    So getting back to a time when radiation balances baryonic matter just requires a redshift of 8100! That sounds kind of reasonable.
    But I still wonder if it is right.

    Right or wrong the prof wont like my method, which uses the radiation constant.

    this radiation constant has an interesting formula in terms of k, hbar, and c. It is

    [tex]\frac{1}{15} \frac {k^4}{(\hbar c)^3}[/tex]

    So if you go to google and put "k^4/(hbar*c)^3/15" in the window, google will give you the value
    that I said earlier-----the 7.566 etc etc.
    Last edited: Mar 19, 2008
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